# Why not just use Bell state entanglement in the CHSH game?

In the CHSH game, both Alice and Bob receive random bits $$x$$ and $$y$$ from a referee Charlie. Based on the bit values and a strategy discussed between Alice and Bob beforehand they will respond with bit values $$a$$ and $$b$$. During the game, Alice and Bob cannot communicate. The goal of the game is to produce matching bits $$a$$ and $$b$$.

The best possible classical strategy is for both Alice and Bob to always respond with a 0, which leads to a $$3/4$$ success probability.

In the quantum case, Alice and Bob share an entangle qubit in the state $$\psi = 1/\sqrt{2}(|0_A0_B\rangle + |1_A1_B\rangle)$$. When Alice receives a bit $$x = 0$$ she measures in the $$|0\rangle, |1\rangle$$ basis and if she gets $$x = 1$$ she measures in the $$|+\rangle, |-\rangle$$ basis. Correspondingly, if Bob receives $$y = 0$$ he measure in $$|a_0\rangle, |a_1\rangle$$, where $$a_0 = \cos(\pi/8)|0\rangle + \sin(\pi/8)|1\rangle$$. If he receives $$y = 1$$ he measure in $$|b_0\rangle, |b_1\rangle$$, where $$b_0 = -\sin(\pi/8)|0\rangle + \cos(\pi/8)|1\rangle$$

I understand the math behind the CHSH game and how the success probability increases to $$\cos(\pi/8)^2$$. However, Alice and Bob have a Bell pair in the state $$\psi = 1/\sqrt{2}(|00\rangle + |11\rangle)$$. Doesn't this mean that if Alice measures $$|0\rangle$$ so will Bob, and similar for $$|1\rangle$$. I believe techniques like quantum teleportation are based on this principle.

Why don't they just agree on a strategy to report back to the referee whatever they have measured? This would lead to a $$100\%$$ success rate. I must be missing something.

• It's quite unclear what you're asking. Can you include some more details? Jan 15 at 1:37
• Thanks @Rammus, I edited the question. Jan 17 at 20:11

"The goal of the game is to produce matching bits $$a$$ and $$b$$."